The corner point method

Step 1: draw the feasible region and find all its corners, by inspection or by solving the two boundary equations that meet at each.

Step 2: evaluate Z at every corner. Call the largest value M and the smallest m.

Step 3: if the region is bounded, M is the maximum and m the minimum of Z.

Dealer: Z = 250x + 75y is 0 at O, 5000 at A(20, 0), 6250 at B(10, 50) and 4500 at C(0, 60). The maximum profit Rs 6250 comes from buying 10 tables and 50 chairs.

B is where 5x + y = 100 meets x + y = 60: subtracting gives 4x = 40, so x = 10 and y = 50.

Maximise Z = 4x + y with x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0: the corners (0, 0), (30, 0), (20, 30), (0, 50) give 0, 120, 110, 50, so the maximum is 120 at (30, 0).

Minimise Z = 200x + 500y with x + 2y ≥ 10, 3x + 4y ≤ 24, x, y ≥ 0: the region is the triangle (0, 5), (4, 3), (0, 6), with Z = 2500, 2300, 3000, so the minimum is 2300 at (4, 3).

Minimising uses exactly the same table; only the choice of the smallest entry changes.